已知函数f(x)=2sin(x+5π/24)cos(x+5π/24)-2cos²(x+5π/24)+1
1个回答
f(x)=sin(2x+5π/12)-cos(2x+5π/12)
=√2sin(2x+5π/12-π/4)
=√2sin(2x+π/6)
所以,最小正周期T=2π/2=π
递增区间:
-π/2+2kπ
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